Question

A population is normally distributed with mean,?=300and ?=20

a.	Find the probability that a value randomly selected from this population will have a value greater than 310
b.	Find the probability that a value randomly selected from this population will have a value less than 285
c.	Find the probability that a value randomly selected from this population will have a value between 285 and 310

Click the icon to view the standard normal table.

Answer 1

Solution:

Given that,

= 300

= 20

a ) p ( x > 310 )

= 1 - p (x < 310 )

= 1 - p ( x -  / ) < ( 310 - 300 /20 )

= 1 - p ( z < 10 / 20 )

= 1 - p ( z < 0.5)

Using z table

= 1 - 0.6915

= 0.3085

Probability = 0.3085

b ) p (x < 285 )

= p ( x -  / ) < ( 285 - 300 /20 )

= p ( z < -15 / 20 )

= p ( z < - 0.75)

Using z table

= 0.2266

Probability = 0.2266

c ) p (285 < x < 310 )

= p ( 285 - 300 /20 ) < ( x -  / ) < ( 310 - 300 /20 )

= p (-15 /20 < z < 10 / 20 )

= p (- 0.75 < z < 0.5)

= p ( z < 0.5) - p p ( z < 0.75 )

Using z table

= 0.6915 - 0.2266

= 0.4649

Probability = 0.4649